A body is projected vertically upwards with a velocity of $19.6 m / s$ . The total time for which the body will remain in the air is (Take $g = 9.8 m / s^{2}$ )

4 s

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6 s

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9 s

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12 s

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Solution

The correct option is A $4 s$
$u = 19.6, a = - 9.8$
using : $s = u t + \frac{a t^{2}}{2}$
when the body reaches ground displacement , $s = 0$
$u t - \frac{g t^{2}}{2} = 0$
$t (2 u - g t) = 0$
we get $t = 0, \frac{2 u}{g}$
since at $t = 0$ body is projected so Time of flight $= \frac{2 u}{g}$
$T = \frac{2 \times 19.6}{9.8}$
$T = 4 s$

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A body is projected vertically upwards with a velocity of 19.6 m/s. The total time for which the body will remain in the air is (Take g=9.8m/s2)

The correct option is A 4 su=19.6,a=−9.8using : s=ut+at22when the body reaches ground displacement ,s=0ut−gt22=0t(2u−gt)=0we get t=0,2ugsince at t=0 body is projected so Time of flight=2ugT=2×19.69.8T=4s

A body is projected vertically upwards with a velocity of $19.6 m / s$ . The total time for which the body will remain in the air is (Take $g = 9.8 m / s^{2}$ )